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Question

A body of mass 2 kg is projected with a speed 4 m/s at an angle 45 to the horizontal. What will be its angular momentum about the point of projection when the body is at the highest point in its trajectory?


A
1.62 kg m2/s
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B
32.62 kg m2/s
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C
162 kg m2/s
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D
3.262 kg m2/s
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Solution

The correct option is D 3.262 kg m2/s

For given projectile motion,
Range (R)=u2sin2θg=42sin(2×45)9.8=169.8 m
Maximum height (Hmax)=u2sin2θ2g=49.8 m
So at maximum height,
r=x ^i+y ^j=R2 ^i+Hmax ^j =(89.8 ^i+49.8 ^j) m
and we know that at highest point velocity will be in x direction only,
u=ux ^i+uy ^j=u cosθ ^i+0 ^j =4 cos45 ^i+0 ^j=(42 ^i+0 ^j) m/s
Now we know,
Angular momentum about origin (L)=r×p
=r×(mu)=m(r×u)=2[(89.8 ^i+49.8 ^j)×(42 ^i+0 ^j)]=2×4×49.8×2(^j×^i)
=3.262(^k) as, (^j×^i=^k)
|L|=3.262 kg m2/s
Hence, option (d) is the correct answer.

Alternate solution:
We know that when the body is at the top most point:
vy=0 ,
vx=vocosθ=4cos45=42m/s
Angular momentum is given by:
L=mass×velocity×perpendicular distance of velocity from the origin
Here perpendicular distance is: Hmax=u2sin2θ2g=49.8 m
Therefore, L=2×42×49.8
L=3.262kg m2/s

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